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2 years ago

Question :

2 answers:

riadik2000 [5.3K]2 years ago

8 0

The second law of motion states that: the acceleration of an object is dependent upon two variables: - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object

<h3>Meaning of Motion</h3>

Motion can be defined as the process of changing position willingly or due to a force applied.

Motion can be seen in different forms and types depending on the object.

In conclusion, The second law of motion is used to deduce the formula for acceleration.

Learn more about second law of motion: brainly.com/question/2009830

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snow_lady [41]2 years ago

4 0

Answer:

Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

Explanation:

Hope it's helpful to you

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4 years ago

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What is the change all living things undergo as the grow

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4 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

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4 years ago

A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0

Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

$M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\$

If we carry out summation of Torques on the pulley we will have to,

$F_2*d-F_1*d = I \alpha$

Where,

I = Inertia moment

$\alpha =$ Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

$I = \frac{1}{2} mr^2$

Replacing this equations we have know that

$(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})$

$F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}$

$F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1$

$F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}$

Replacing our values we have that

$F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}$

$F_1 = 0.974 N$

Therefore the tension in the string between the pulley and the cart is 0.974 N

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4 years ago

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bearhunter [10]

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3 years ago

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